MATH 340 LAB 2 Assignment

MATH 340
LAB 2 Assignment

Root-finding: Bisection Method
As outlined in your textbook, the bisection method serves to find a zero of a realvalued function f(x). The first thing to keep in mind is that we want to solve an
equation of the form f(x) = 0 for x. Therefore if the function is not given in that
form, you should recast it to have everything on one side. For instance, if you are
asked to find the root of cos(x) = x, you should write it as f(x) = cos(x) − x in
order to solve the equation f(x) = 0 for x.
This being said, your book uses the notation [a, b] for the interval in which we
are finding the root. Call this root r, such that f(r) = 0. The main assumption
under which the method works is that the function has a change of sign at the
endpoints of the interval, therefore a zero in between. This is translated in the
condition f(a)*f(b)<0.
A pseudo-code to implement this method is briefly described by the following
steps:
Given the initial interval [a, b] s. t. f(a)f(b) < 0
while (b-a)/2 > tol
1. Define c = (a + b)/2. % this is our approximation to the root r
2. if f(c) == 0, stop, end % in this case we are lucky, we’ve already found r
3. if f(a)f(c)<0,
4. b=c
5. else
6. a=c, end
1
end % end of while loop
Again, update the approximate solution c = (a + b)/2.
Problem
Write your own code for the bisection method. In addition to what the pseudocode provided above does, it is worthwhile to count how many steps it takes to
converge to the solution (hint: to keep track, increase a counter variable n in the
loop, with the instruction n = n+1). At each step, print out the interval endpoints,
the midpoint, and function values at the endpoints and the approximate root (the
midpoint). Also, print out a message to say whether you have found the solution
according to the condition in the step no. 2, or your best approximation after the
loop has ended.
Now use your code to solve the problems below. For each one write a sentence
or two conclusion as appropriate. For example,“My code found the root to be
. . . after . . . steps to within an error of . . . in x and the function value at this value
of x is . . . This seems to be as expected as the error theory implies it should take
about . . . steps”. Of course, if something different occurs, youd want to highlight
that aspect.
1. Find an interval of length 1 with integer endpoints on which a solution to
x
3 = 9 lies. Then use your bisection code starting with this interval to find
the solution to within 10−4
.
2. Find an interval of length 1 with integer endpoints on which a solution to
6 + cos2 x = x lies. Then use your bisection code starting with this interval
to find the solution to within 10−4
.
3. Suppose we want to know what interest rate we would need to achieve (assuming constant interest rates) such that the initial deposit of $50, 000 and
annual contribution of $10, 000 will grow to $1, 000, 000 in 20 years. Thus
we need to solve 1000000 =
50000 + 10000
r


e
20r −
10000
r
for the interest rate,
r. Find a reasonable interval on which the solution ought to lie and use the
bisection code to find r to within 10−6
.